Find the value of - 1 - 3 i 8 + - 1 + 3 i 8 .

We have - 1 - 3 i 8 + - 1 + 3 i 8 = 2 - 1 2 - 3 i 2 8 + 2 - 1 2 + 3 i 2 8 We know that - 1 2 ± 3 i 2 are the values of the imaginary cube roots of unity, which are ω and ω 2 . Also, we know that 1 + ω + ω 2 = 0 , ω 3 n = 1 , ω 3 n + 1 = ω and ω 3 n + 2 = ω 2 . Hence, if ω = - 1 2 - 3 i 2 then ω 2 = - 1 2 + 3 i 2 . Thus, 2 - 1 2 - 3 i 2 8 + 2 - 1 2 + 3 i 2 8 = 2 8 ω 8 + 2 8 ω 2 8 = 256 ω 8 + ω 16 = 256 ω 3 × 2 + 2 + ω 3 × 5 + 1 = 256 ω 2 + ω = 256 - 1 = - 256 .

Let z be a complex number which satisfy z - 1 - i = 2 . Then find the locus of z .

Let z = x + i y be a complex number which satisfy z - 1 - i = 2 . Then, we have x + i y - 1 - i = 2 ⇒ x - 1 + i y - 1 = 2 Using a + i b = a 2 + b 2 , ⇒ x - 1 2 + y - 1 2 = 2 ⇒ x - 1 2 + y - 1 2 = 2 . The above equation is in the form x - h 2 + y - k 2 = r 2 , which is the equation of a circle with centre h , k and radius r . Hence, the equation x - 1 2 + y - 1 2 = 2 represents a circle with centre 1 , 1 and radius 2 . Thus, the locus of z is a circle.

Let z be a non-zero complex number, if I m 1 z is zero, then find the locus of z .

Let z = x + i y . Then 1 z = 1 x + i y ⇒ 1 z = 1 x + i y × x - i y x - i y ⇒ 1 z = x - i y x 2 - i 2 y 2 ⇒ 1 z = x - i y x 2 + y 2 , ∵ i 2 = - 1 . Given I m 1 z = 0 ⇒ - y x 2 + y 2 = 0 ⇒ y = 0 . Thus z = x , which is a purely real number. Thus, the locus of z is a line.

E M B I B E

Gamma Question Bank for Engineering Mathematics>Class XI>Chapter 5 - Complex Numbers>Assignment

Embibe Experts Solutions for Exercise 7: Assignment

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Exercise 7: Assignment

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